# Check if a large number can be divided into two or more segments of equal sum

Given a very large Number N. The task is to check if the number can be divided into two or more segments of an equal sum.

Examples:

```Input: N = 73452
Output: Yes
Segments of {7}, {3, 4}, {5, 2} which has equal sum of 7

Input: N = 1248
Output: No
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The following steps can be followed to solve the problem:

1. Since the number can be large, the number is initialized in a string.
2. Use prefixSum array to store the prefix sum of the array.
3. Now traverse from second element to last, and the first segment thus will be 0 to i-1, whose sum is Prefixsum[i-1].
4. Use another pointer which traverses from i to n, and keep adding the sum.
5. If the sum at any stage is equal to Prefixsum[i-1], then the segment has a sum equal to first.
6. Reinitialize the segment sum value to 0 and keep moving the pointer.
7. If at any stage the segment sum exceeds the sum of the first segment, then break, as the division with segment sum as prefixsum[i-1] is not possible.
8. If the pointer reaches the last number, check if the last segment sum is equal to the first segment sum i.e., prefixsum[i-1], then it can be divided into segments of equal sum.

Below is the implementation of the above approach.

## C++

 `// C++ program to Check if a large number can be divided  ` `// into two or more segments of equal sum ` `#include ` `using` `namespace` `std; ` ` `  `// Function to check if a number ` `// can be divided into segments ` `bool` `check(string s) ` `{ ` `    ``// length of string ` `    ``int` `n = s.length(); ` ` `  `    ``// array to store prefix sum ` `    ``int` `Presum[n]; ` ` `  `    ``// first index ` `    ``Presum = s - ``'0'``; ` ` `  `    ``// calculate the prefix ` `    ``for` `(``int` `i = 1; i < n; i++) { ` `        ``Presum[i] = Presum[i - 1] + (s[i] - ``'0'``); ` `    ``} ` ` `  `    ``// iterate for all number from second number ` `    ``for` `(``int` `i = 1; i <= n - 1; i++) { ` ` `  `        ``// sum from 0th index to i-1th index ` `        ``int` `sum = Presum[i - 1]; ` `        ``int` `presum = 0; ` `        ``int` `it = i; ` ` `  `        ``// counter turns true when sum ` `        ``// is obtained from a segment ` `        ``int` `flag = 0; ` ` `  `        ``// iterate till the last number ` `        ``while` `(it < n) { ` `            ``// sum of segments ` `            ``presum += s[it] - ``'0'``; ` ` `  `            ``// if segment sum is equal ` `            ``// to first segment ` `            ``if` `(presum == sum) { ` `                ``presum = 0; ` `                ``flag = 1; ` `            ``} ` `            ``// when greater than not possible ` `            ``else` `if` `(presum > sum) { ` `                ``break``; ` `            ``} ` `            ``it++; ` `        ``} ` ` `  `        ``// if at the end all values are traversed ` `        ``// and all segments have sum equal to first segment ` `        ``// then it is possible ` `        ``if` `(presum == 0 && it == n && flag == 1) { ` `            ``return` `true``; ` `        ``} ` `    ``} ` `    ``return` `false``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string s = ``"73452"``; ` `    ``if` `(check(s)) ` `        ``cout << ``"Yes"``; ` `    ``else` `        ``cout << ``"No"``; ` `    ``return` `0; ` `} `

## Java

 `// Java program to Check if a large number can be divided  ` `// into two or more segments of equal sum ` ` `  `public` `class` `GFG { ` ` `  `    ``// Function to check if a number ` `    ``// can be divided into segments ` `    ``static` `boolean` `check(String s) ` `    ``{ ` `        ``// length of string ` `        ``int` `n = s.length(); ` `      `  `        ``// array to store prefix sum ` `        ``int``[] Presum = ``new` `int``[n]; ` `      `  `        ``// first index ` `        ``char``[] s1 = s.toCharArray(); ` `        ``Presum[``0``] = s1[``0``] - ``'0'``; ` `      `  `        ``// calculate the prefix ` `        ``for` `(``int` `i = ``1``; i < n; i++) { ` `            ``Presum[i] = Presum[i - ``1``] + (s1[i] - ``'0'``); ` `        ``} ` `      `  `        ``// iterate for all number from second number ` `        ``for` `(``int` `i = ``1``; i <= n - ``1``; i++) { ` `      `  `            ``// sum from 0th index to i-1th index ` `            ``int` `sum = Presum[i - ``1``]; ` `            ``int` `presum = ``0``; ` `            ``int` `it = i; ` `      `  `            ``// counter turns true when sum ` `            ``// is obtained from a segment ` `            ``int` `flag = ``0``; ` `      `  `            ``// iterate till the last number ` `            ``while` `(it < n) { ` `                ``// sum of segments ` `                ``presum += s1[it] - ``'0'``; ` `      `  `                ``// if segment sum is equal ` `                ``// to first segment ` `                ``if` `(presum == sum) { ` `                    ``presum = ``0``; ` `                    ``flag = ``1``; ` `                ``} ` `                ``// when greater than not possible ` `                ``else` `if` `(presum > sum) { ` `                    ``break``; ` `                ``} ` `                ``it++; ` `            ``} ` `      `  `            ``// if at the end all values are traversed ` `            ``// and all segments have sum equal to first segment ` `            ``// then it is possible ` `            ``if` `(presum == ``0` `&& it == n && flag == ``1``) { ` `                ``return` `true``; ` `            ``} ` `        ``} ` `        ``return` `false``; ` `    ``} ` `      `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) { ` `         `  `        ``String s = ``"73452"``; ` `        ``if` `(check(s)) ` `            ``System.out.println(``"Yes"``); ` `        ``else` `            ``System.out.println(``"No"``); ` `    ``} ` `} `

## Python 3

 `# Python program to Check if a large number can be divided  ` `# into two or more segments of equal sum ` `# Function to check if a number  ` `# can be divided into segments  ` `def` `check(s) : ` ` `  `    ``# length of string  ` `    ``n ``=` `len``(s) ` ` `  `    ``# array to store prefix sum  ` `    ``Presum ``=` `[``0``] ``*` `n ` ` `  `    ``# ord() gives ASCII value ` `    ``# first index  ` `    ``Presum[``0``] ``=` `ord``(s[``0``]) ``-` `ord``(``'0'``) ` ` `  `    ``# calculate the prefix  ` `    ``for` `i ``in` `range``(n) : ` `        ``Presum[i] ``=` `Presum[i ``-` `1``] ``+` `(``ord``(s[i]) ``-` `ord``(``'0'``)) ` ` `  `    ``# iterate for all number from second number ` `    ``for` `i ``in` `range``(n) : ` ` `  `        ``# sum from 0th index to i-1th index  ` `        ``sum` `=` `Presum[i] ` `        ``presum ``=` `0` `        ``it ``=` `1` ` `  `        ``#  counter turns true when sum  ` `        ``# is obtained from a segment ` `        ``flag ``=` `0` ` `  `        ``# iterate till the last number ` `        ``while``( it < n) : ` ` `  `            ``# sum of segments                          ` `            ``presum ``+``=` `ord``(s[it]) ``-` `ord``(``'0'``)                              ` ` `  `            ``# if segment sum is equal  ` `            ``# to first segment  ` `            ``if` `presum ``=``=` `sum` `: ` `                ``presum ``=` `0` `                ``flag ``=` `1` ` `  `            ``# when greater than not possible                         ` `            ``elif` `presum > ``sum` `:                 ` `                ``break` `                                      `  `            ``it ``+``=` `1` ` `  `        ``# if at the end all values are traversed  ` `        ``# and all segments have sum equal to first segment  ` `        ``# then it is possible  ` `        ``if` `presum ``=``=` `0` `and` `it ``=``=` `n ``and` `flag ``=``=` `1` `: ` `            ``return` `True` ` `  `    ``return` `False` ` `  `                                      `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``s ``=` `"73452"` `    ``if` `check(s) : ` `        ``print``(``"Yes"``) ` `    ``else` `: ` `        ``print``(``"No"``) ` ` `  `# This code is contributed by ANKITRAI1 `

## C#

 `// C# program to Check if a large ` `// number can be divided into two ` `// or more segments of equal sum  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` ` `  `    ``// Function to check if a number  ` `    ``// can be divided into segments  ` `    ``static` `bool` `check(String s)  ` `    ``{  ` `        ``// length of string  ` `        ``int` `n = s.Length;  ` `     `  `        ``// array to store prefix sum  ` `        ``int``[] Presum = ``new` `int``[n];  ` `     `  `        ``// first index  ` `        ``char``[] s1 = s.ToCharArray();  ` `        ``Presum = s1 - ``'0'``;  ` `     `  `        ``// calculate the prefix  ` `        ``for` `(``int` `i = 1; i < n; i++)  ` `        ``{  ` `            ``Presum[i] = Presum[i - 1] + (s1[i] - ``'0'``);  ` `        ``}  ` `     `  `        ``// iterate for all number from second number  ` `        ``for` `(``int` `i = 1; i <= n - 1; i++)  ` `        ``{  ` `     `  `            ``// sum from 0th index to i-1th index  ` `            ``int` `sum = Presum[i - 1];  ` `            ``int` `presum = 0;  ` `            ``int` `it = i;  ` `     `  `            ``// counter turns true when sum  ` `            ``// is obtained from a segment  ` `            ``int` `flag = 0;  ` `     `  `            ``// iterate till the last number  ` `            ``while` `(it < n)  ` `            ``{  ` `                ``// sum of segments  ` `                ``presum += s1[it] - ``'0'``;  ` `     `  `                ``// if segment sum is equal  ` `                ``// to first segment  ` `                ``if` `(presum == sum) ` `                ``{  ` `                    ``presum = 0;  ` `                    ``flag = 1;  ` `                ``}  ` `                 `  `                ``// when greater than not possible  ` `                ``else` `if` `(presum > sum) ` `                ``{  ` `                    ``break``;  ` `                ``}  ` `                ``it++;  ` `            ``}  ` `     `  `            ``// if at the end all values are traversed  ` `            ``// and all segments have sum equal to first segment  ` `            ``// then it is possible  ` `            ``if` `(presum == 0 && it == n && flag == 1) ` `            ``{  ` `                ``return` `true``;  ` `            ``}  ` `        ``}  ` `        ``return` `false``;  ` `    ``}  ` `     `  `    ``// Driver Code  ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{  ` `         `  `        ``String s = ``"73452"``;  ` `        ``if` `(check(s))  ` `            ``Console.WriteLine(``"Yes"``);  ` `        ``else` `            ``Console.WriteLine(``"No"``);  ` `    ``}  ` `}  ` ` `  `// This code has been contributed by 29AjayKumar `

Output:

```Yes
```

Time Complexity: O(N2)
Auxiliary Space: O(N)

My Personal Notes arrow_drop_up Striver(underscore)79 at Codechef and codeforces D

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